Size of sets associated to Gaussian integers Given a non-zero Gaussian integer $z$, we define the set $\mathcal S(z)$
containing all solutions
of $ab+cd=z$ satisfying $\min(\vert a\vert,\vert b\vert)>\max(\vert c\vert,\vert d\vert)$ with $a,b,c,d$ in the set $\mathbb Z[i]$ of Gaussian integers.
The identity
$$2m+1=(2n+(2n^2-m-1)i)(2n-(2n^2-m-1)i)-(2n^2-m)^2$$
and a similar identity for even integers implies that $\mathcal S(z)$ is infinite for every ordinary non-zero integer.
More generally $\mathcal S(z)$ is therefore infinite if $z$ is of the form $i^k s^2m$ for $s$ a non-zero Gaussian integer and for $m$
a non-zero ordinary integer.
Is there a computable bound on the size of solutions if
$\mathcal S(z)$ is finite?
(The set $\mathcal S(1+i)$ for example seems to be finite).
 A: 1. Let me address the case of $z=1+i$. It seems to me that those are the only counterexamples, but a proof I can imagine is now hidden under technical details.
Consider the representation $z=ab+cd$ and define integers $\alpha=|a|^2$, $\beta=|b|^2$, $\gamma=|c|^2$, $\delta=|d|^2$. WLOG we have
$$
  \alpha\geq \beta>\gamma\geq\delta.
$$
Let the vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$ represent $z$ (fixed) and $ab$ (variable), so that the vector $\overrightarrow{BA}$ represents $cd$. The sides of $OAB$ are $\sqrt2$, $\sqrt{\alpha\beta}$ and $\sqrt{\gamma\delta}$.
Assume that $\alpha\geq \gamma+2$. Then we have
$$
  \sqrt2 \geq \sqrt{\alpha\beta}-\sqrt{\gamma\delta}\geq \sqrt{(\gamma+1)(\gamma+2)}-\gamma,
$$
so
$$
  (\gamma+1)(\gamma+2)\leq (\gamma+\sqrt2)^2,
$$
which provides an explicit upper bound on $\gamma$ (and then a bound on $\alpha$ is also easy).
A similar argument works if $\delta\leq \beta-2$.
It remains to check the case $\alpha=\beta=\gamma+1=\delta+1$. Let $BH$ be an altitude of the triangle. Then $H$ represents an integer multiple of $\frac{1+i}2$, so $AH$ is an integer multiple of $\frac1{\sqrt2}$. However,
$$
  2AH\pm\sqrt2=OH+AH=\frac{OH^2-AH^2}{OH-AH}=\frac{OB^2-AB^2}{\sqrt2}=\pm\frac{2\gamma+1}{\sqrt2},
$$
so that $AH$ is an odd multiple of $\frac1{2\sqrt2}$. This contradiction finishes the proof.
2. Some remarks on what to do in the general case.
The first part can be applied similarly, yielding that, if $S(z)$ is infinite, then there are infinitely many solutions with small and fixed values of $\alpha-\beta$, $\beta-\gamma$, and $\gamma-\delta$.
But it seems that one such series should exist. This boils down to some equations similar in vein to Pell equations, which I have no time to dig through right now, sorry.
